3.117 \(\int \cos ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=127 \[ -\frac{8 a^2 \cos ^5(c+d x)}{33 d \sqrt{a \sin (c+d x)+a}}-\frac{64 a^3 \cos ^5(c+d x)}{231 d (a \sin (c+d x)+a)^{3/2}}-\frac{256 a^4 \cos ^5(c+d x)}{1155 d (a \sin (c+d x)+a)^{5/2}}-\frac{2 a \cos ^5(c+d x) \sqrt{a \sin (c+d x)+a}}{11 d} \]

[Out]

(-256*a^4*Cos[c + d*x]^5)/(1155*d*(a + a*Sin[c + d*x])^(5/2)) - (64*a^3*Cos[c + d*x]^5)/(231*d*(a + a*Sin[c +
d*x])^(3/2)) - (8*a^2*Cos[c + d*x]^5)/(33*d*Sqrt[a + a*Sin[c + d*x]]) - (2*a*Cos[c + d*x]^5*Sqrt[a + a*Sin[c +
d*x]])/(11*d)

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Rubi [A]  time = 0.234739, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2674, 2673} \[ -\frac{8 a^2 \cos ^5(c+d x)}{33 d \sqrt{a \sin (c+d x)+a}}-\frac{64 a^3 \cos ^5(c+d x)}{231 d (a \sin (c+d x)+a)^{3/2}}-\frac{256 a^4 \cos ^5(c+d x)}{1155 d (a \sin (c+d x)+a)^{5/2}}-\frac{2 a \cos ^5(c+d x) \sqrt{a \sin (c+d x)+a}}{11 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-256*a^4*Cos[c + d*x]^5)/(1155*d*(a + a*Sin[c + d*x])^(5/2)) - (64*a^3*Cos[c + d*x]^5)/(231*d*(a + a*Sin[c +
d*x])^(3/2)) - (8*a^2*Cos[c + d*x]^5)/(33*d*Sqrt[a + a*Sin[c + d*x]]) - (2*a*Cos[c + d*x]^5*Sqrt[a + a*Sin[c +
d*x]])/(11*d)

Rule 2674

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

Rule 2673

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m - 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rubi steps

\begin{align*} \int \cos ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=-\frac{2 a \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{11 d}+\frac{1}{11} (12 a) \int \cos ^4(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{8 a^2 \cos ^5(c+d x)}{33 d \sqrt{a+a \sin (c+d x)}}-\frac{2 a \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{11 d}+\frac{1}{33} \left (32 a^2\right ) \int \frac{\cos ^4(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=-\frac{64 a^3 \cos ^5(c+d x)}{231 d (a+a \sin (c+d x))^{3/2}}-\frac{8 a^2 \cos ^5(c+d x)}{33 d \sqrt{a+a \sin (c+d x)}}-\frac{2 a \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{11 d}+\frac{1}{231} \left (128 a^3\right ) \int \frac{\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac{256 a^4 \cos ^5(c+d x)}{1155 d (a+a \sin (c+d x))^{5/2}}-\frac{64 a^3 \cos ^5(c+d x)}{231 d (a+a \sin (c+d x))^{3/2}}-\frac{8 a^2 \cos ^5(c+d x)}{33 d \sqrt{a+a \sin (c+d x)}}-\frac{2 a \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{11 d}\\ \end{align*}

Mathematica [A]  time = 0.154282, size = 69, normalized size = 0.54 \[ -\frac{2 \left (105 \sin ^3(c+d x)+455 \sin ^2(c+d x)+755 \sin (c+d x)+533\right ) \cos ^5(c+d x) (a (\sin (c+d x)+1))^{3/2}}{1155 d (\sin (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-2*Cos[c + d*x]^5*(a*(1 + Sin[c + d*x]))^(3/2)*(533 + 755*Sin[c + d*x] + 455*Sin[c + d*x]^2 + 105*Sin[c + d*x
]^3))/(1155*d*(1 + Sin[c + d*x])^4)

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Maple [A]  time = 0.113, size = 77, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ){a}^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3} \left ( 105\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+455\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+755\,\sin \left ( dx+c \right ) +533 \right ) }{1155\,d\cos \left ( dx+c \right ) }{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+a*sin(d*x+c))^(3/2),x)

[Out]

2/1155*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)^3*(105*sin(d*x+c)^3+455*sin(d*x+c)^2+755*sin(d*x+c)+533)/cos(d*x+c)/(
a+a*sin(d*x+c))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cos \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^(3/2)*cos(d*x + c)^4, x)

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Fricas [A]  time = 1.68689, size = 471, normalized size = 3.71 \begin{align*} -\frac{2 \,{\left (105 \, a \cos \left (d x + c\right )^{6} + 245 \, a \cos \left (d x + c\right )^{5} - 20 \, a \cos \left (d x + c\right )^{4} + 32 \, a \cos \left (d x + c\right )^{3} - 64 \, a \cos \left (d x + c\right )^{2} + 256 \, a \cos \left (d x + c\right ) +{\left (105 \, a \cos \left (d x + c\right )^{5} - 140 \, a \cos \left (d x + c\right )^{4} - 160 \, a \cos \left (d x + c\right )^{3} - 192 \, a \cos \left (d x + c\right )^{2} - 256 \, a \cos \left (d x + c\right ) - 512 \, a\right )} \sin \left (d x + c\right ) + 512 \, a\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{1155 \,{\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-2/1155*(105*a*cos(d*x + c)^6 + 245*a*cos(d*x + c)^5 - 20*a*cos(d*x + c)^4 + 32*a*cos(d*x + c)^3 - 64*a*cos(d*
x + c)^2 + 256*a*cos(d*x + c) + (105*a*cos(d*x + c)^5 - 140*a*cos(d*x + c)^4 - 160*a*cos(d*x + c)^3 - 192*a*co
s(d*x + c)^2 - 256*a*cos(d*x + c) - 512*a)*sin(d*x + c) + 512*a)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*
sin(d*x + c) + d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(a+a*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cos \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^(3/2)*cos(d*x + c)^4, x)